1. The bombardier beetle uses an explosive discharge as a defensive measure. The chemical reaction involved is the oxidation of hydroquinone by hydrogen peroxide to produce quinine and water:

C6H4(OH)2 (aq) + H2O2 (aq) �� C6H4O2 (aq) + 2 H2O (l)
Calculate ΔH for this reaction from the following data:
C6H4(OH)2 aq --> C6H4O2 aq + H2 g ΔH = +177.4 kJ
H2 g + O2 g --> H2O2 aq ΔH = -191.2 kJ
H2 g + ½ O2 g --> H2O g ΔH = -241.8 kJ
H2O g --> H2O (L) ΔH = -43.8 kJ

Add equation 1 to the reverse of equation 2, then add twice equation 3 and twice equation 4. Cancel atoms/molecules common to both sides and check to make sure you have the equation you want. Don't forget to change the sign of any delta H value for which the equation was reversed. Also, don't forget to multiply the delta H value for any equation multiplied.

-202.6 kJ

To calculate the enthalpy change (ΔH) for the given chemical reaction, we can use Hess's law. Hess's law states that the overall enthalpy change of a reaction is the sum of the enthalpy changes of the individual reactions that make up the overall reaction.

In this case, we need to manipulate the given chemical reactions and combine them in a way that cancels out some of the reactants or products. Let's start by rearranging the given reactions to match the reactants and products in the target reaction:

1. C6H4(OH)2(aq) --> C6H4O2(aq) + H2(g) (ΔH = +177.4 kJ)
2. H2(g) + O2(g) --> H2O2(aq) (ΔH = -191.2 kJ)
3. H2(g) + 1/2 O2(g) --> H2O(g) (ΔH = -241.8 kJ)
4. H2O(g) --> H2O(L) (ΔH = -43.8 kJ)

Now, let's reverse reaction 1 and multiply it by a factor of 2 so that the hydroquinone (C6H4(OH)2) cancels out:

5. 2C6H4O2(aq) + 2H2(g) --> 2C6H4(OH)2(aq) (ΔH = -2 * (+177.4 kJ) = -354.8 kJ)

Next, let's reverse reaction 2 and multiply it by a factor of 2 so that the hydrogen peroxide (H2O2) cancels out:

6. 2H2O2(aq) --> 2H2(g) + O2(g) (ΔH = 2 * (-191.2 kJ) = -382.4 kJ)

Now, let's reverse reaction 3 and multiply it by a factor of 2 so that the water (H2O) cancels out:

7. 2H2O(g) --> 2H2(g) + O2(g) (ΔH = 2 * (-241.8 kJ) = -483.6 kJ)

Finally, let's sum up reactions 4, 5, 6, and 7 to obtain the target reaction:

8. 2C6H4O2(aq) + 2H2O(g) --> 2C6H4(OH)2(aq) + O2(g) (ΔH = -354.8 kJ + (-483.6 kJ) + (-43.8 kJ) = -882.2 kJ)

So, the enthalpy change (ΔH) for the given reaction is -882.2 kJ.

Is it -194.6KJ ?